Proposition 7

Let E be a separated locally convex space and let $u$ be an endomorphism of E.

a) Let $E_1$ be a closed subspace of finite codimension of E, stable under $u$. Let $u_1$ denote the element of $\mathscr{L}(E_1)$ which coincides with $u$ on $E_1$. Then $u$ is a Riesz endomorphism if and only if $u_1$ is one;

b) Suppose that $u$ is a Fredholm endomorphism of index 0. Let $\widehat{E}$ be the completion of E and let $\widehat{u}\in \mathscr{L}(\widehat{E})$ be the extension of $u$ by continuity. Then $u$ is a Riesz endomorphism if and only if $\widehat{u}$ is one;

c) Endow the dual $E'$ of E with the topology of bounded convergence, the topology of compact convergence, or the weak topology. If $u$ is a Riesz endomorphism of E, then $^tu$ is a Riesz endomorphism of $E'$.

a) In order that $u$ be a Fredholm endomorphism of index zero, it is necessary and sufficient that $u_1$ be one (III, p. 44, prop. 3). On the other hand, for every $n\in \mathbf{N}$, one has Ker($u^n_1$) $= E_1\cap$ Ker($u^n$), whence

dim Ker($u^n_1$)$\leqslant$ dim Ker($u^n$)$\leqslant$ dim Ker($u^n_1$) $+$ codim$_E(E_1)$,

therefore the sequence (dim Ker($u^n$))$_{n\in\mathbf{N}}$ is bounded if and only if the sequence (dim Ker($u^n_1$))$_{n\in\mathbf{N}}$ is. Hence, $u$ is a Riesz endomorphism of E if and only if $u_1$ is a Riesz endomorphism of $E_1$ (lemma 2).

b) By prop. 4 of III, p. 44, for every $n\in \mathbf{N}$, the mapping $\widehat{u}^n$ is a Fredholm endomorphism of $\widehat{E}$ such that Ker($\widehat{u}^n$) $=$ Ker($u^n$) and ind($\widehat{u}^n$) $=$ ind($u^n$) $=n$ ind($u$). Hence, $u$ is a Riesz endomorphism if and only if $\widehat{u}$ is a Riesz endomorphism.

c) The transpose $^tu$ is a Fredholm endomorphism of index 0 of $E'($III, p. 43, n$^o3$). Since the kernel of $(^tu)^n$ is the orthogonal of the image of $u^n$ (EVT, IV, p. 27, prop. 2), the sequence (Ker($^tu$)$^n$)$_n$ is stationary. This proves that $^tu$ is a Riesz endomorphism of $E'$.